2
\$\begingroup\$

I am having two boards(1.5Feet x 1Feet each) which are connected in a backplane. I2C master is in BOARD1 and the slave(Temperature Sensor) in BOARD2. In 100KHz mode, I am able to read the slave properly but when I move to 400KHz mode I am not gettign any response from slave.

Please find the attached diagram for the I2C diagram, there are some MUX and SWITCHES are connected on the I2C lines. I have shown only SCL in the picture but SDA is also the same connection.

In the above case there are 3 pull up resistors on both SCL and SDA line. Can anyone tell us how to find the right pull up value for all three resistors.

And one more thing is BOARD1 and BOARD2 has their own 3.3V supply generated on board. I have also written the input capacitance of each devices in the picture.

\$\endgroup\$
  • \$\begingroup\$ The schematic didn't come through. Upload your schematic somewhere (pdf or picture) and add a link to your post. \$\endgroup\$ – Nick Alexeev Jan 3 '13 at 7:34
  • \$\begingroup\$ Please add schematics and datasheet links for the I2C parts. Also, a brief explanation of what the boards are expected to do, may help. \$\endgroup\$ – Anindo Ghosh Jan 3 '13 at 7:35
6
\$\begingroup\$

Since your I2C works at 100Khz, but not 400 Khz, it is a good idea to look at the various factors that have an effect on timing.

1: Check that your slave board supports 400Khz.

2: Resistor values are too big.

When the timing is increased from 100k to 400k, the period of the clock drops from 10 us to 2.5 us.

This means that the rising edge of your data/clock signals has a significantly less amount of time to settle. the time taken is calculated as follows:

t = rc

the capacitance on the bus is usually constant and a property of each device. It sounds like you have these. Add them up.

The resistor values are the next variable. Since you have three in parallel, you need to add them using 1/Rt = 1/R1 + 1/R2 + 1/R3 and so on. You only need one resistor on the bus, so having three in parallel is going to lower the total resistance.

You can now calculate t using the above formula. If it is more than 300ns (just over 10% of your clock period at 400k), then the rise time is out of I2C spec. Here, table 5, page 32.

If you'd like to calculate the correct resistor value, you can re-arrange the above formula to get R=t/c and work from there, where T is 300ns or less.

\$\endgroup\$
  • \$\begingroup\$ Nice answer. But there is also a minimum boundary for the resistor value. If it is really low, the rise time will be very small, which is nice. In fact not that nice if your product has to comply with some EMC standard. But if the value if too small, the drivers will drain more current to pull the lines low. At the limit, the drivers would not by able to pull the lines below the Vil thresholds. \$\endgroup\$ – Blup1980 Jan 3 '13 at 10:15
  • \$\begingroup\$ I knew there was a minimum, but I didn't know the exact reasoning offhand. haven't encountered it often enough! on that thought though, the high current would cause the edge to overshoot and cause nasty ringing, messing up the emc? \$\endgroup\$ – stanri Jan 3 '13 at 10:58
  • \$\begingroup\$ The quicker the signal (short rise or fall time) the more harmonics you have in it. And you really dislikes theses harmonics when doing EMC compliance. \$\endgroup\$ – Blup1980 Jan 5 '13 at 8:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.